Requirements of a probability distribution:1.2.Section 4.3The mean or Expected Valueof a discrete random variable x is: Example: How much money on average will an insurance company make off of a 1-year life insurance policy worth $10,000, if they charge $290 for the policy, and you have a 0.999 probability of surviving the year? *Note even if the company has to pay out it keeps the $290 for itself, so a payout is only $10,000 - $290 = $9,710.Example: What is your expected value on the following game? You offer your friends $10 if they can roll a six on a die, but they pay a dollar amount equal to what they roll on the die if they roll any number other than 6. Should your friends play this game against you?The population variance for a random variablex is (29(2922222( )( )E xxp xxp xσμμμ=-=-=-����The population standard deviation for a random variable xis given by taking the square root of the above mentioned population variance: Example: The following table gives the probability that x patients out of five will be cured. Calculate the expected value for the r.v. x, calculate the standard deviation, and use Chebyshev’s rule to interpret the standard deviation. Section 4.4Characteristics of a binomial random variable:1.Experiment consists of n identical trials2.There are only two possible outcomes for each trial (success or failure)3.The probability of a success remains the same from trial to trial4.The trials are independent5.The random variable is the number of successes in n trialsExample: Is the following a binomial experiment? A marketing firm conducts a survey to determine if consumers prefer the appearance of the bottle for Absolute Vodka over the bottle of its two top rivals. 1000 people will be asked to pick their favorite bottle, and the number of people who select the Absolute bottle will be counted. What if we instead recorded the name of the brand that was chosen by each person?11

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Example: Is the following a Binomial experiment? Let x represent the number of correct guesses on 5 multiple choice questions where each question has 5 answer options. Example: Derive the probability distribution for the above problemX012345P(x)The Binomial Probability Distribution( )xnxnp xp qx-��=����for x = 0, 1, 2, . . . , np = probability of a successq = 1 - px = number of successesn = number of trialsMean, Variance, and Standard deviation for a Binomial Random Variable:Mean = n pμVariance = 2n p qσ=��Standard Deviation = n p qσ=��Example: Find the mean and standard deviation for the number of correct guesses on the above mentioned five-question multiple choice quiz. Would it be unusual to pass a five-question quiz by blind guessing?=